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Orlov [11]
3 years ago
10

What is the solution to y=-x+3 and y=4x-2

Mathematics
1 answer:
DIA [1.3K]3 years ago
3 0

Answer: x=1

              y=2

Step-by-step explanation:

y=-x+3

y=4x-2

-----------

    y=y

-x+3=4x-2

3+2=4x+x

5=5x

x=1

y=-1+3

y=2

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5 0
3 years ago
Confused on factoring
Dmitriy789 [7]

Answer:

The answer I got was 16

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Step-by-step explanation:

3 0
3 years ago
) a, p and d are n×n matrices. check the true statements below:
BigorU [14]
A. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).

b. True. Suppose \mathbf P is the matrix of the eigenvectors of \mathbf A, and \mathbf D is the diagonal matrix of the eigenvalues of \mathbf A:


\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}

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Then

\mathbf{AP}=\begin{bmatrix}\mathbf{Av}_1&\cdots&\mathbf{Av}_n\end{bmatrix}=\begin{bmatrix}\lambda_1\mathbf v_1&\cdots&\lambda_n\mathbf v_n\end{bmatrix}=\mathbf{PD}

In other words, the columns of \mathbf{AP} are \mathbf{Av}_i, which are identically \lambda_i\mathbf v_i, and these are the columns of \mathbf{PD}.

c. False. A counterexample is the matrix

\begin{bmatrix}1&1\\0&1\end{bmatrix}

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

\begin{bmatrix}0&1\\0&0\end{bmatrix}

with eigenvalue \lambda=0 and eigenvector \begin{bmatrix}k&0\end{bmatrix}^\top, where k\in\mathbb R. But the matrix can't be diagonalized.
7 0
3 years ago
Wat is -27/3 wil mark brainless
KATRIN_1 [288]

Answer:

=9

Step-by-step explanation:

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3 0
3 years ago
Someone please help me will give BRAILIEST!!!!!
mel-nik [20]
Pretty sure the answer is 1.
Hope this helps. 
(One point up from 1998, then 1 to the right gets you to 2000. 1/1 = 1)
5 0
3 years ago
Read 2 more answers
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